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Choose the correct option.
∫₀^(2a) f(x) dx = ∫₀^a f(x) dx + ∫₀^a f(k-x) dx, then the value of k is:
$\int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(k-x) \, dx$, then the value of $k$ is:
- (a)$a$
- (b)$2a$
- (c)Independent of $a$
- (d)None of the above options
Answer
Answer
Correct option: (b)
Answer: (b) $2a$
$\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_a^{2a} f(x)\,dx$.
In the second integral put $x = 2a - t$: it becomes $\int_0^a f(2a-t)\,dt$, so $k = 2a$.